Physics for ME · Module 6 of 16
Work, Energy, and Power
Energy bookkeeping answers questions that force-by-force analysis makes painful: how far, how fast, how much fuel, how big a motor.
Readiness check
From Modules 3 to 5 and Math Module 5. Tick only what you can do closed-notes.
- Compute net force from an FBD.
- Use v² = v₀² + 2as confidently.
- Evaluate ∫F dx for a varying force (Math Module 5).
- Keep joules, watts, and newton-metres distinct.
- Project a force onto a displacement direction (dot product).
The core idea
Work moves energy between accounts. The totals always balance.
W = F·d (along motion)KE = ½mv²P = W/t = F·vThe work-energy theorem says net work equals the change in kinetic energy. With potential energy (mgh, ½ks²) on the books, conservation turns many force problems into one-line audits: energy in, energy out, energy stored.
The skills, taught in order
Energy methods trade force-by-force analysis for accounting. Learn what each account holds and how work moves energy between them, and many problems collapse to one balanced line.
6.1 Work by a constant force
Work is force times the distance moved along the force: W = Fs cos φ, where φ is the angle between the force and the displacement. Only the along-motion component counts; a force at right angles (a normal force, a centripetal pull) does no work. Work is measured in joules (1 J = 1 N·m) and may be positive, negative, or zero.
Micro-example. Pushing a crate 4 m with 50 N at 30° above the motion does W = 50 × 4 × cos 30° = 173 J.
6.2 Work by a varying force
When the force changes along the path, the work is the area under the force-distance curve, W = ∫F dx. Stretching an ideal spring by x stores W = ½kx², the triangle under the line F = kx.
6.3 Kinetic energy and the work-energy theorem
A moving mass carries kinetic energy KE = ½mv². The work-energy theorem says the total work by all forces equals the change in kinetic energy: Wtotal = ½mv₂² − ½mv₁². It answers "what speed?" by accounting, with no need for time.
6.4 Potential energy
Some work is stored and recoverable. Lifting a mass stores gravitational PE = mgh; compressing or stretching a spring stores elastic PE = ½ks². Both are energy waiting to return to motion.
6.5 Conservation of energy
If only conservative forces (gravity, springs) act, mechanical energy is constant: KE₁ + PE₁ = KE₂ + PE₂. That one line replaces a force-by-force solution for many problems, and only the endpoints matter, not the path between them.
Micro-example. A mass dropped from height h lands at √(2gh), because mgh becomes ½mv² regardless of the route down.
6.6 Friction as the leak
Friction does negative work and turns mechanical energy into heat that never returns. Carry it as a loss term: KE₁ + PE₁ = KE₂ + PE₂ + (friction heat). It is the everyday face of the second law you meet in Thermodynamics.
| Account | Formula | Recoverable? |
|---|---|---|
| Kinetic | ½mv² | yes |
| Gravitational PE | mgh | yes |
| Elastic (spring) PE | ½ks² | yes |
| Friction heat | f·d | no, one way |
6.7 Power
Power is the rate of doing work, P = W/t on average and P = F·v at an instant. Its unit is the watt (1 W = 1 J/s); one horsepower is 746 W. Power, not energy, sizes a motor: the same job done faster needs more of it.
Micro-example. Lifting 200 kg at 0.5 m/s needs P = Fv = (200 × 9.81)(0.5) = 981 W, about 1.3 hp before losses.
6.8 Efficiency
No real machine delivers all the energy it draws. Efficiency η is useful output over total input; the remainder leaves as heat, noise, or friction. Size a drive above the ideal figure by dividing by η.
| Quantity | Formula | Unit |
|---|---|---|
| Work | W = Fs cos φ | J = N·m |
| Kinetic energy | ½mv² | J |
| Power (average) | P = W/t | W = J/s |
| Power (instant) | P = Fv | W |
Engineering connection: Dynamics, Thermodynamics (the energy balance of Module 12), machine and motor sizing, and Energy Systems.
Worked example 1: braking distance by energy audit
A 1200 kg car at 25 m/s (90 km/h) brakes with a steady total force of 7200 N. Find the braking distance, then state what happens at double the speed.
- ProblemFind the stopping distance in Figure 1.
- Given / findm = 1200 kg, v = 25 m/s, F = 7200 N steady. Find d; then d at 50 m/s.
- AssumptionsLevel road, constant braking force, all kinetic energy into brake heat.
- ModelWork-energy theorem: the brakes must do work equal to the kinetic energy.
- Equations½mv² = F·d
- SolveKE = 0.5 × 1200 × 625 = 375 000 J. d = 375 000/7200 = 52.1 m. At 50 m/s: KE quadruples (1.5 MJ), so d = 208 m.
- CheckForce route: a = 7200/1200 = 6 m/s²; d = v²/2a = 625/12 = 52.1 m. Same answer, two methods. The v² scaling matches Module 1's module.
- ConclusionDouble the speed, four times the distance: the single most consequential equation in road safety, and the same audit sizes brakes, crash barriers, and flywheel absorbers.
Worked example 2: a spring launcher
A spring of stiffness k = 1200 N/m is compressed 0.15 m and launches a 0.5 kg ball straight up. Find the speed at which the ball leaves the spring and the height it reaches, by moving energy between the spring, kinetic, and gravitational accounts.
- ProblemFind the launch speed and the peak height for the spring launcher in Figure 2.
- Given / findk = 1200 N/m, compression s = 0.15 m, m = 0.5 kg, g = 9.81 m/s². Find v at release and the height H above the start.
- AssumptionsIdeal spring, no air drag, vertical motion, the spring is massless so all its stored energy goes to the ball.
- ModelConservation of energy. Stored spring energy ½ks² converts to kinetic energy and gravitational potential energy.
- Equations½ks² = ½mv² + mgs (at release) ½ks² = mgH (at the peak)
- SolveStored energy ½ks² = ½(1200)(0.15²) = 13.5 J. At release: ½mv² = 13.5 − (0.5)(9.81)(0.15) = 12.76 J, so v = √(2 × 12.76 / 0.5) = 7.15 m/s. At the peak: H = 13.5 / (0.5 × 9.81) = 2.75 m above the start.
- CheckThe same 13.5 J appears in all three accounts in turn, never gaining or losing total. Dropping the ball from 2.75 m would return it to 7.3 m/s at the start level, recovering the launch energy minus the small spring-extension rise.
- ConclusionOne conserved total carried the ball through a spring, a flight, and a stop, with no force or time needed. Springs storing and returning energy this way are the heart of suspensions, valve trains, and the oscillations of Module 10.
Misconceptions and diagnostics
| Mistake | Symptom | Diagnostic question | Correction |
|---|---|---|---|
| Perpendicular forces credited with work | Normal force or centripetal force "doing work" | "Does the force have a component along the motion?" | W = Fd cos θ: perpendicular means cos 90° = 0. No work. |
| Friction losses recovered | Energy audits that balance after a skid | "Which account did the friction energy land in?" | Heat. It never returns to the mechanical books. Write it as a loss term. |
| Energy and power interchanged | Motors sized in joules, batteries in watts | "Is this an amount or a rate?" | Energy (J) is the amount; power (W = J/s) is the rate. P = Fv links them. |
| KE linear in speed | "Twice the speed, twice the energy" | "What power of v sits in ½mv²?" | Squared. Double speed means quadruple energy and quadruple braking distance. |
Practice ladder
A hoist lifts a 200 kg pallet 6 m at constant speed in 8 s. Find the work done and the power required.
Show answer
W = mgh = 200 × 9.81 × 6 = 11.77 kJ. P = 11 772/8 = 1.47 kW. Add 20 to 30% for a real motor's losses.
A 2 kg slider is released from rest at the top of a frictionless ramp 1.8 m high. Find its speed at the bottom, and explain why the ramp angle does not matter.
Show answer
mgh = ½mv²: v = √(2 × 9.81 × 1.8) = 5.94 m/s. Gravity's work depends only on the height drop, so every frictionless path from that height gives the same speed.
The Module 4 sled (net force 48.1 N on 20 kg) starts from rest. Use energy methods to find its speed after 10 m, and check with kinematics.
Show answer
Net work = 48.1 × 10 = 481 J = ½ × 20 × v², so v = √48.1 = 6.94 m/s. Kinematics: v = √(2 × 2.41 × 10) = 6.94 m/s. The theorem is Newton plus kinematics, pre-integrated.
Size a motor for a real task you choose (a winch, a conveyor lift, an e-bike hill climb): state mass, height or force, time target, and efficiency, and produce the required power with a margin.
What good work looks like
A clean energy audit, P = W/t with units, an efficiency assumption cited, and a sensible standard motor size chosen above the computed need.
Working with AI, and proving it yourself
Use AI as an examiner, not a solver
Portfolio task
Write a one-page energy audit of a real system you use (your commute, a kettle, a gym session): all accounts, all transfers, the losses named, and one efficiency number computed from your own estimates.
Retrieval and spaced review
Closed notes. Answer out loud, then reveal.
1. Define work, with the angle dependence.
W = Fd cos θ: force times displacement times alignment. Perpendicular force does none.
2. State the work-energy theorem.
Net work on a body equals its change in kinetic energy: ΣW = Δ(½mv²).
3. Write the two potential energies of mechanics.
Gravitational mgh, spring ½ks². Both are stored work, recoverable.
4. What makes friction special in energy accounting?
Its work converts mechanical energy to heat irreversibly: a one-way leak, the seed of the second law in Thermodynamics.
5. Two ways to compute power?
P = W/t for averages and P = F·v instantaneously: the second sizes drives at speed.